Inital acceleration to free-fall

AI Thread Summary
A model rocket accelerates upward at 12 m/s² until it reaches a height of 26 m, after which it enters free fall. The initial velocity at this height is calculated to be 25 m/s. The total flight duration, including ascent and descent, is approximately 5.98 seconds, with the maximum height attained being around 100 m. The speed of the rocket just before impact is determined to be -33.7 m/s, indicating downward motion. The discussion emphasizes the importance of applying the equations of motion correctly to solve for each phase of the rocket's flight.
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Homework Statement



A model rocket blasts off and moves upward with an acceleration of 12 m/s^2 until it reaches a height of 26 m, at which point the engine shuts off and it continues its flight in free fall. A)what is the maximum height attained by the rocket? B) What is the speed of the rocket just before it hits the ground? C) What is the total duration of the rocket's flight?

Homework Equations



In the Example that is somewhat like this question in the book is using the equation:
Xf= Xi + Vi(t) - 0.5(g)t^2 = 0,
but this makes absolutely no sense since the intial acceleration is given as 12 m/s^2

The Attempt at a Solution


I have attemped this question and several others in different ways and I just can not find a equation that I can manipulate so it fits into a free fall situation. Do I just assume the 12 m/s^2 for the first 26 m and then from there incorporate the acceleration of gravity 9.81 m/s^2 is it coming straight back down or is it forming a parabola? It is just way too confusing, and 4 hours at this problem is just way too much. Any help or suggestions would be greatly appreciated!
 
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consider this as a vertical motion problem
All the relevant equations are the 3 equations of motion.

first you know rocket starts with a given accel and also the initial velocity is 0. Find final velocity at 26m height by one of the 3 equations of motion. From then on it's just freefall.
But remember that since the rocket has certain velocity it goes up for some time and then comes down. so include that time in the whole motion.
 
Thank you for the help I am not sure if I did it right, but here it goes:

I first solved for v= 25 m/s

Then I used the quadratic formula to find the time of the flight = 5.98 sec

Certain velocity = 4.41 m/s and added to 25m/s so I could solve for max height.

Then I solved for max height = 100m

Lastly I solved for the speed of the rocket before it hit the ground = -33.7 (- indicating direction.

* Like I said I think I did it wrong so, any corrections are appreciated
 
I believe you are wrong. the mistake begins from time of flight. Try stickin to the equations of motion and you will find it much easier. :smile:
 
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